Body A of mass $$4$$ m moving with speed u collides with another body B of mass $$2$$ m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is

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Infinity Learn · Accepted Answer

According to conservation of momentum,$$4mu1=4mv1+2mv2$$⇒ $$2u1-v1=v2$$                          .........(i) From conversation of energy, $$12(4m)u12=12(4m)v12+12(2m)v22$$ ⇒ $$2u12-v12=v22$$                          ........(ii) From (i) and (ii) $$2u12-v12=4u1-v12$$                    ..........(iii)  Now, fraction of loss in kinetic energy for mass $$4m$$ ,  Δ$$KKi=Ki-KfKi=1/2(4m)u12-12(4m)v1212(4m)u12$$             .........(iv)  Substituting (iii) in (iv), we get Δ$$KKi=89$$

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Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is

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